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Let f be a bounded real valued function on [a,b], then define the length of the curve L(f) as follows: \begin{equation}\tag{1} L(f)=sup\sum\limits_{i=1}^{n}\sqrt{ (x_i-x_{i-1})^2+(f(x_i)-f(x_{i-1}))^2}, \end{equation} where sup is taken over all partitions of [a,b].

The question is to show, L(f) is equal to \begin{equation}\tag{2} \lim\limits_{max|x_i-x_{i-1}|\to 0}\sum\limits_{i=1}^{n}\sqrt {(x_i-x_{i-1})^2+(f(x_i)-f(x_{i-1}))^2} \end{equation} for continuous functions. I know when $max|x_i-x_{i-1}|\to 0$, we get finer partitions so that $(1)\leq(2)$ because if we pick any partition and then let $max|x_i-x_{i-1}|\to0$ then we get a greater sum due to triangle inequality.(I guess what I said is true and does not require continuity.) I could not show $(2)\leq(1)$ and could not use the continuity. In a finite sum, we can switch $\sum$ and limit but as $max|x_i-x_{i-1}|\to0$ we get an infinite sum so switching may fail. I got stuck.

P.S. I have read (Determining the Length of a Curve Using Partitions) but I did not understand "there is a sequence $(Q_n)_{n\geq0}$ of partitions with (3) holds" which was stated in the answer. If you refer to that question, please tell me why (3) holds. I thought that (3) is the same as what we want to show in that question.

Probably, I missed an easy detail. Thanks in advance for any help.

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Sedergine
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  • Can't you just say that the partitions in the limit are also part of the set for which you look at the supremum, and therefore, they are always less then or equal to the $\sup$? – Ragnar Dec 19 '13 at 23:53
  • I thought what you said, but then I still did not use continuity. Apart from that, for example the sequence 1/n is always greater than 0 but if we take the limit we don't get a positive number; in short, I am not sure we are in part of the set in the limiting process without using continuity. – Sedergine Dec 20 '13 at 00:03

2 Answers2

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The basic fact you are (or should be) using is that if $S$ is any set of real numbers, and $X=x_1, \dotsc, x_n, \dotsc$ is any convergent subsequence, then $L=\lim X \leq \sup S.$ This is obvious, because if $L > \sup S,$ then there is an $i$ such that $x_i \geq L - (L-\sup S),$ contradicting the supremum property.

Igor Rivin
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  • As you have mentioned, with this basic property we get equality, but where we use continuity then. – Sedergine Dec 20 '13 at 00:15
  • Probably, you meant that in order to guarantee the existence of the limit we need continuity, but I could not see why this needs to be true formally; even if intuitively this sounds valid. – Sedergine Dec 20 '13 at 00:25
  • Well, if $f$ is a discontinuous (but everywhere defined) function, you will get a limit, but it won't be the length of the curve (you will be adding in the jumps, just consider $f(x) =0$ for $x<1,$ and $f(x)=1$ for $x>1.$ The points $(0, 0)$ and $(2, 1)$ lie on the curve, but the limit you get is something different). – Igor Rivin Dec 20 '13 at 00:46
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Continuity is necessary, because otherwise, the 'length' of a curve would not be defined well. You could get an infinite number of jumps from $0$ to $1$ etc, so the supremum could be infinite.

Ragnar
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    The OP is allowing the sup to be infinite. However, if you think about what is being computed, it makes no sense for a discontinuous map. – Igor Rivin Dec 20 '13 at 00:10